Integrand size = 28, antiderivative size = 57 \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=-\frac {F^{c (a+b x)}}{e (d+e x)}+\frac {b c F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log (F)}{e^2} \] Output:
-F^(c*(b*x+a))/e/(e*x+d)+b*c*F^(c*(a-b*d/e))*Ei(b*c*(e*x+d)*ln(F)/e)*ln(F) /e^2
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\frac {F^{a c} \left (-\frac {e F^{b c x}}{d+e x}+b c F^{-\frac {b c d}{e}} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log (F)\right )}{e^2} \] Input:
Integrate[F^(c*(a + b*x))/(d^2 + 2*d*e*x + e^2*x^2),x]
Output:
(F^(a*c)*(-((e*F^(b*c*x))/(d + e*x)) + (b*c*ExpIntegralEi[(b*c*(d + e*x)*L og[F])/e]*Log[F])/F^((b*c*d)/e)))/e^2
Time = 0.39 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2007, 2608, 2609}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {F^{c (a+b x)}}{(d+e x)^2}dx\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {b c \log (F) \int \frac {F^{c (a+b x)}}{d+e x}dx}{e}-\frac {F^{c (a+b x)}}{e (d+e x)}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {b c \log (F) F^{c \left (a-\frac {b d}{e}\right )} \operatorname {ExpIntegralEi}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{e^2}-\frac {F^{c (a+b x)}}{e (d+e x)}\) |
Input:
Int[F^(c*(a + b*x))/(d^2 + 2*d*e*x + e^2*x^2),x]
Output:
-(F^(c*(a + b*x))/(e*(d + e*x))) + (b*c*F^(c*(a - (b*d)/e))*ExpIntegralEi[ (b*c*(d + e*x)*Log[F])/e]*Log[F])/e^2
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m _), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))) , x] - Simp[f*g*n*(Log[F]/(d*(m + 1))) Int[(c + d*x)^(m + 1)*(b*F^(g*(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && In tegerQ[2*m] && !TrueQ[$UseGamma]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.74
method | result | size |
risch | \(-\frac {\ln \left (F \right ) b c \,F^{b c x} F^{a c}}{e^{2} \left (b c x \ln \left (F \right )+\frac {b c \ln \left (F \right ) d}{e}\right )}-\frac {\ln \left (F \right ) b c \,F^{\frac {c \left (e a -b d \right )}{e}} \operatorname {expIntegral}_{1}\left (-b c x \ln \left (F \right )-a c \ln \left (F \right )-\frac {-e a \ln \left (F \right ) c +b c d \ln \left (F \right )}{e}\right )}{e^{2}}\) | \(99\) |
Input:
int(F^(c*(b*x+a))/(e^2*x^2+2*d*e*x+d^2),x,method=_RETURNVERBOSE)
Output:
-ln(F)*b*c/e^2*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+b*c*ln(F)/e*d)-ln(F)*b*c/e^2 *F^(c*(a*e-b*d)/e)*Ei(1,-b*c*x*ln(F)-a*c*ln(F)-(-e*a*ln(F)*c+b*c*d*ln(F))/ e)
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35 \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=-\frac {F^{b c x + a c} e - \frac {{\left (b c e x + b c d\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right ) \log \left (F\right )}{F^{\frac {b c d - a c e}{e}}}}{e^{3} x + d e^{2}} \] Input:
integrate(F^((b*x+a)*c)/(e^2*x^2+2*d*e*x+d^2),x, algorithm="fricas")
Output:
-(F^(b*c*x + a*c)*e - (b*c*e*x + b*c*d)*Ei((b*c*e*x + b*c*d)*log(F)/e)*log (F)/F^((b*c*d - a*c*e)/e))/(e^3*x + d*e^2)
\[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\int \frac {F^{c \left (a + b x\right )}}{\left (d + e x\right )^{2}}\, dx \] Input:
integrate(F**((b*x+a)*c)/(e**2*x**2+2*d*e*x+d**2),x)
Output:
Integral(F**(c*(a + b*x))/(d + e*x)**2, x)
\[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{e^{2} x^{2} + 2 \, d e x + d^{2}} \,d x } \] Input:
integrate(F^((b*x+a)*c)/(e^2*x^2+2*d*e*x+d^2),x, algorithm="maxima")
Output:
integrate(F^((b*x + a)*c)/(e^2*x^2 + 2*d*e*x + d^2), x)
\[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{e^{2} x^{2} + 2 \, d e x + d^{2}} \,d x } \] Input:
integrate(F^((b*x+a)*c)/(e^2*x^2+2*d*e*x+d^2),x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)/(e^2*x^2 + 2*d*e*x + d^2), x)
Timed out. \[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{d^2+2\,d\,e\,x+e^2\,x^2} \,d x \] Input:
int(F^(c*(a + b*x))/(d^2 + e^2*x^2 + 2*d*e*x),x)
Output:
int(F^(c*(a + b*x))/(d^2 + e^2*x^2 + 2*d*e*x), x)
\[ \int \frac {F^{c (a+b x)}}{d^2+2 d e x+e^2 x^2} \, dx=f^{a c} \left (\int \frac {f^{b c x}}{e^{2} x^{2}+2 d e x +d^{2}}d x \right ) \] Input:
int(F^((b*x+a)*c)/(e^2*x^2+2*d*e*x+d^2),x)
Output:
f**(a*c)*int(f**(b*c*x)/(d**2 + 2*d*e*x + e**2*x**2),x)